A) x - 7 > 3 |add 7 to both sides
x > 10
b) x - 7 > 3 |add 7 to both sides
x > 10
4sin²(x) = 5 - 4cos(x)
4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
2 - 2cos(2x) = 5 - 4cos(x)
- 2 - 2
-2cos(2x) = 3 - 4cos(x)
-2[2cos²(x) - 1] = 3 - 4cos(x)
-4cos²(x) + 2 = 3 - 4cos(x)
- 2 - 2
-4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
4cos²(x) - 4cos(x) + 1 = 0
[2cos(x) - 1]² = 0
2cos(x) - 1 = 0
+ 1 + 1
2cos(x) = 1
2 2
cos(x) = ¹/₂
cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
x = 60, 300
x = π/3, 5π/3
[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π
Answer: The percent of red lights last between 2.5 and 3.5 minutes = 95.44% approx.
Step-by-step explanation:
Given: Mean time = 3 minutes
standard deviation = 0.25 minutes
The probability of red lights last between 2.5 and 3.5 minutes :-
The percent of red lights last between 2.5 and 3.5 minutes = 95.44% approx.
Answer:
postulates of 1-3 : if two independent planes intersect , they intersect in exactly 1 line . hope its helpful .
